TITLE: Principal Ideal Groupring
QUESTION [5 upvotes]: Let $R[G]$ be a groupring (not necessarily commutative). Under which conditions on $R$ and $G$ is $R[G]$ a Principal Ideal Ring, respectively a Principal Ideal Domain (not commutative either)?

REPLY [3 votes]: You can see at the wiki article that the question of detecting zero divisors in general group rings is not a trivial task. You will also find citations there for special cases of $G$ and $R$ that do make $R[G]$ a domain.
It looks like there is a characterization for Artinian principal ideal rings appearing in this paper:

Dorsey, Thomas J.(1-CCR)
  Morphic and principal-ideal group rings. (English summary) 
  J. Algebra 318 (2007), no. 1, 393–411. 
  16S34 

The paper also contains references to and an appendix concerning a more classical result due to the work of Sehgal, Fisher and Passman which says that if $R$ is a division ring and $G$ is a nilpotent group, then $R[G]$ is a principal ideal ring. (Of course this is a special case of the main theorem of the paper, but I thought you might be interested.)
It's tempting to suspect that the question is open in the general (non-Artinian) case. PIR's are of course Noetherian, and we don't even know for which groups $G$ it happens that $R[G]$ is Noetherian. 

Another interesting thing I recently learned is that $|H|^{-1}\sum_{g\in H}g$ is an idempotent of $R[G]$ whenever $H$ is a finite subgroup of $G$ such that $|H|$ is a unit in $R[G]$. This means if $R$ has characteristic zero, $|G|>1$, and $R[G]$ is a domain, then $G$ can't have any finite subgroups other than $\{1\}$.